Correct Answer - Option 3 : 0 & 0
Concept:
1) For unit step input u(t), the steady state error is given by:
\({e_{ss}} = \frac{1}{{1 + {K_p}}}\)
\({K_p} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
G(s) H(s) = open loop transfer function of the system
2) For unit ramp input r(t) = tu(t), the steady state error is given by:
\({e_{ss}} = \frac{1}{{{K_v}}}\)
\({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
3) For parabolic input \(p\left( t \right) = \frac{{{t^2}}}{2}u\left( t \right)\), the steady-state error is:
\({e_{ss}} = \frac{1}{{{K_a}}}\)
\({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Example:
Consider a type-II system as:
\(G\left( s \right)H\left( s \right) = \frac{K}{{{s^2}\left( {s + a} \right)}}\)
\({K_p} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
\(= \mathop {\lim }\limits_{s \to 0} \frac{K}{{{s^2}\left( {s + a} \right)}} = \infty \)
\({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
\(= \mathop {\lim }\limits_{s \to 0} \frac{{sK}}{{{s^2}\left( {s + a} \right)}}\)
\(= \mathop {\lim }\limits_{s \to a} \frac{K}{{s\left( {s + a} \right)}} = \infty \)
ess for the unit step will be:
\({e_{ss}} = \frac{1}{{1 + {K_p}}} = \frac{1}{{1 + \infty }} = 0\)
ess for unit ramp will be:
\({e_{ss}} = \frac{1}{{{K_v}}} = \frac{1}{\infty } = 0\)
